So I was checking the structure of nitric acid in Wikipedia, however I couldn't quite fathom why it looked like that because it seemed to contradict the following statement:

A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.

enter image description here

If we draw it like the one on the right, we get rid of formal charges and the structure is said to be more "stable". Why does this concept not work in this case?

  • $\begingroup$ The left one satisfies 8-electron rule (something like that...) $\endgroup$
    – user26143
    Dec 2, 2013 at 17:56
  • 1
    $\begingroup$ Yeah but let's take a look at another example: H3PO4 Although the octet formation is not fulfilled, the structure with no formal charges is preferred. Here is the comparison: i.imgur.com/XLpgIjn.png $\endgroup$ Dec 2, 2013 at 18:03
  • $\begingroup$ How do you know which one is prefered? $\endgroup$
    – user26143
    Dec 2, 2013 at 18:04
  • 6
    $\begingroup$ It is quite strict that the number of electrons around N should not exceed 8. Having lower than 8 electron is not ideal but still OK. Nitrogen has only 4 valance orbital, enough to hold 8 electrons. It is not a matter of preference-- the second structure is simply impossible. It is just like you cannot draw a double bond to a hydrogen atom. $\endgroup$ Dec 2, 2013 at 18:16
  • 2
    $\begingroup$ It is very important to know that Lewis octet rule is one thing and the Lewis Formula another. That does not mean that they may not be combined. The main flaw of Lewis formula is that it can only describe covalent bonds with two electrons each. Hence a Lewis structure is never a complete representation of the bonding picture. $\endgroup$ Apr 15, 2014 at 5:19

5 Answers 5


Nitrogen is in the second row with no $d$ orbital in the valence shell. It obeys the octet rule and cannot have more than 8 electron.

There are exceptions to the octet rule. Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.

As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule. However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.

According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts. However, their contributions are not even. some of them are more favorable than others.

For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds. Based on octet rule alone, there are 3 possible resonance structures that are favorable.

Resonance structures of HNO3

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges. As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.

It is also common to write a mixture of resonance structures as the hybrid form

Hybrid form of HNO3

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms. The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.


Well! here is a short quick reply..!!

I was pretty sure that the lewis stucture of $\ce{HNO3}$ would be the one with 0 formal charge and than I found this. Though this link only shows how to calculate formal charge on $\ce{HNO3}$ but it gives a little hint.

Secondly a thought came to my mind that if you will focus on Resonance, there will be double bond character on both bonds between N & O.



Historically, there was nothing wrong with structures like the one you drew on the right. Iwan Ostromisslensky had no problems drawing 4-chloro-1,2-dinitrobenzene with a pentavalent nitrogen in 1908.[1] (Yes, a nitro group is not nitrate or nitric acid, but after some time of searching I just took what I found to prove my point.)

pentavalent nitrogen

Staudinger and Meyer drew dimethylanilinoxide similarly in 1919:[2]

dimethylanilin oxide

I wasn’t able to find the point in time when preferences changed so I gave up. It must have been sometime around when the quantum chemistry of orbitals was becoming better and better understood, and it was realised that there are only four orbitals nitrogen has access to for bonding (2s and three 2p). Henceforth, people described structures with nitrogen in the centre with four bonds only, better reflecting reality.

For elements of higher periods, most notably sulfur and phosphorus, many people still choose to draw too many bonds rather than write charge-separated structures. This is often explained with ‘d-orbital participation’ — but from a practical point of view, the 3d orbital has a very similar energy to 4s yet nobody suggests 4s participation. All ‘extended octet’ structures can be drawn in a way that conforms to the octet rule, so maybe it is only a matter of time until $\ce{P=O}$ bonds in phosphate disappear.

As for rules to determine the likelihood of Lewis structures, this is a better set:

  1. Lewis structures in which all atoms have an octet (doublet for hydrogen) are preferred.

  2. If 1. cannot be fulfilled, Lewis structures which have the least number of elements with sub-octet structures are preferred.

  3. If 1. or 2. generate a set of possible structures choose one that has a minimal number of formal charges.

  4. If 3. leaves a set of possible structures, choose one where the formal charges are distributed according to electronegativity (electronegative elements having negative formal charges).

  5. If 4. leaves a set of possible structures, choose one in which the formal charges are closer together.

  6. If your final structure has an expanded octet on a main group element, start again at 1.


[1]: I. Ostromisslensky, J. Prakt. Chem. 1908, 78, 263. DOI: 10.1002/prac.19080780121.

[2]: H. Staudinger, J. Meyer, Helv. Chim. Acta 1919, 2, 608. DOI: 10.1002/hlca.19190020161.

  • $\begingroup$ I have even seen trifluoroamine oxide drawn with a double bond from nitrogen to oxygen making 10 electrons around nitrogen. The nitrogen-oxygen bond actually does have double-bond character, but this comes from ionic contributing structures of the form $\ce{(O=NF_2^+)F^-}$, not anything with ten valence electrons on nitrogen. $\endgroup$ Dec 5, 2017 at 18:58

Even though you may have a filled octet, if the central atom has a positive formal charge, we generally form double bonds until the formal charge is reduced to 0 as close as possible as that will be the more stable configuration. However, there are some exceptions.

Please take a look at the chlorate ion. For chlorate, you would expect to leave four single bonded oxygens to the chlorine, however, we are left with a formal charge of +3 on the chlorine and -1 on each of the oxygens. Therefore, we form double bonds until the formal charge is removed and are left with only a formal charge of -1 on the single bonded oxygen.

edit: looking at your question the structure on the left is preferred even though the one on the right has a lower formal charge.

Perhaps Wikipedia has the wrong structure for the Nitric acid page? It looks like all the oxygen atoms are single bonded in the figure to the top right.


Looking for images, Wikipedia does have the right resonance structures with the double bonded oxygen, however, there is only one double bond contrary to what we expect which is two double bonds.


  • $\begingroup$ Yes, indeed. Doesn't this principle contradict with the structure of HNO3 though? The formal charge of the center atom (nitrogen) is said to be +1. $\endgroup$ Dec 2, 2013 at 18:15
  • $\begingroup$ @ZaferCesur Chlorine and nitrogen are fundamentally different. Nitrogen atom has no d orbital because it is the in second period. It has only s and p and the maximum number of electron it can hold is 8. Chlorine is in 3rd period. It can hold up to 18 electrons. $\endgroup$ Dec 2, 2013 at 18:20
  • $\begingroup$ Okay, I got it now! You should maybe post an answer so that I can upvote or something. $\endgroup$ Dec 2, 2013 at 18:24
  • $\begingroup$ @XiaoleiZhu Yep, in this case it is not possible to form two double bonds as extended octets are formed only by atoms with vacant d-orbitals in the valence shell (p-elements from the third or later periods). Thanks for the clarification. $\endgroup$
    – user3735
    Dec 2, 2013 at 18:26
  • 1
    $\begingroup$ The Wiki picture is a graph that is just a bond and angle mapping. Even in chlorine (esp., and lower elements) d-orbitals have no role in bonding (@XiaoleiZhu). The concept of hypervalency (see also goldbook) is still under criticism and one usually refers to four-electron-three-centre bonds, rather than including d-orbitals in main group elements. $\endgroup$ Apr 15, 2014 at 5:12

enter image description hereThree oxygens are bounded to one nitrogen atom.Oxygen being more electronegative attracts the pair of electron more strogly.One of the oxygen binds with an atom of hydrogen and share an electron with nitrogen to complete its octet and the remaining oxygen atom share an other electron with nitrogen atom. In this way electron pairs are partially donated and coordinate covalent or dative bond is formed


Not the answer you're looking for? Browse other questions tagged or ask your own question.